\(\int e^{\coth ^{-1}(x)} x (1+x) \, dx\) [279]

Optimal result

Integrand size = 9, antiderivative size = 99 \[ \int e^{\coth ^{-1}(x)} x (1+x) \, dx=\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}} x+\frac {1}{3} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {\frac {-1+x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3+\text {arctanh}\left (\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right ) \]

[Out]

arctanh((1+1/x)^(1/2)*((-1+x)/x)^(1/2))+1/3*(1+1/x)^(3/2)*x^2*((-1+x)/x)^(1/2)+1/3*(1+1/x)^(5/2)*x^3*((-1+x)/x
)^(1/2)+x*(1+1/x)^(1/2)*((-1+x)/x)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6310, 6315, 98, 96, 94, 212} \[ \int e^{\coth ^{-1}(x)} x (1+x) \, dx=\text {arctanh}\left (\sqrt {\frac {1}{x}+1} \sqrt {\frac {x-1}{x}}\right )+\frac {1}{3} \left (\frac {1}{x}+1\right )^{5/2} \sqrt {\frac {x-1}{x}} x^3+\frac {1}{3} \left (\frac {1}{x}+1\right )^{3/2} \sqrt {\frac {x-1}{x}} x^2+\sqrt {\frac {1}{x}+1} \sqrt {\frac {x-1}{x}} x \]

[In]

Int[E^ArcCoth[x]*x*(1 + x),x]

[Out]

Sqrt[1 + x^(-1)]*Sqrt[(-1 + x)/x]*x + ((1 + x^(-1))^(3/2)*Sqrt[(-1 + x)/x]*x^2)/3 + ((1 + x^(-1))^(5/2)*Sqrt[(
-1 + x)/x]*x^3)/3 + ArcTanh[Sqrt[1 + x^(-1)]*Sqrt[(-1 + x)/x]]

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6310

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*x^p*(1 + c/(d*
x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[n/2] && Inte
gerQ[p]

Rule 6315

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[-c^p, Subst[Int[(1 +
d*(x/c))^p*((1 + x/a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2))), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ
[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int e^{\coth ^{-1}(x)} \left (1+\frac {1}{x}\right ) x^2 \, dx \\ & = -\text {Subst}\left (\int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^4} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3-\frac {2}{3} \text {Subst}\left (\int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^3} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{3} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3-\text {Subst}\left (\int \frac {\sqrt {1+x}}{\sqrt {1-x} x^2} \, dx,x,\frac {1}{x}\right ) \\ & = \sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}} x+\frac {1}{3} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3-\text {Subst}\left (\int \frac {1}{\sqrt {1-x} x \sqrt {1+x}} \, dx,x,\frac {1}{x}\right ) \\ & = \sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}} x+\frac {1}{3} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3+\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right ) \\ & = \sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}} x+\frac {1}{3} \left (1+\frac {1}{x}\right )^{3/2} \sqrt {-\frac {1-x}{x}} x^2+\frac {1}{3} \left (1+\frac {1}{x}\right )^{5/2} \sqrt {\frac {-1+x}{x}} x^3+\text {arctanh}\left (\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.41 \[ \int e^{\coth ^{-1}(x)} x (1+x) \, dx=\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x \left (5+3 x+x^2\right )+\log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \]

[In]

Integrate[E^ArcCoth[x]*x*(1 + x),x]

[Out]

(Sqrt[1 - x^(-2)]*x*(5 + 3*x + x^2))/3 + Log[(1 + Sqrt[1 - x^(-2)])*x]

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.63

method	result	size
trager	\(\frac {\left (1+x \right ) \left (x^{2}+3 x +5\right ) \sqrt {-\frac {1-x}{1+x}}}{3}+\ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )\)	\(62\)
risch	\(\frac {\left (x^{2}+3 x +5\right ) \left (x -1\right )}{3 \sqrt {\frac {x -1}{1+x}}}+\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{\sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\)	\(62\)
default	\(\frac {\left (x -1\right ) \left (\left (\left (x -1\right ) \left (1+x \right )\right )^{\frac {3}{2}}+3 x \sqrt {x^{2}-1}+3 \ln \left (x +\sqrt {x^{2}-1}\right )+6 \sqrt {x^{2}-1}\right )}{3 \sqrt {\frac {x -1}{1+x}}\, \sqrt {\left (x -1\right ) \left (1+x \right )}}\)	\(67\)

[In]

int(1/((x-1)/(1+x))^(1/2)*x*(1+x),x,method=_RETURNVERBOSE)

[Out]

1/3*(1+x)*(x^2+3*x+5)*(-(1-x)/(1+x))^(1/2)+ln((-(1-x)/(1+x))^(1/2)*x+(-(1-x)/(1+x))^(1/2)+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int e^{\coth ^{-1}(x)} x (1+x) \, dx=\frac {1}{3} \, {\left (x^{3} + 4 \, x^{2} + 8 \, x + 5\right )} \sqrt {\frac {x - 1}{x + 1}} + \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*x*(1+x),x, algorithm="fricas")

[Out]

1/3*(x^3 + 4*x^2 + 8*x + 5)*sqrt((x - 1)/(x + 1)) + log(sqrt((x - 1)/(x + 1)) + 1) - log(sqrt((x - 1)/(x + 1))
- 1)

Sympy [F]

\[ \int e^{\coth ^{-1}(x)} x (1+x) \, dx=\int \frac {x \left (x + 1\right )}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \]

[In]

integrate(1/((-1+x)/(1+x))**(1/2)*x*(1+x),x)

[Out]

Integral(x*(x + 1)/sqrt((x - 1)/(x + 1)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.11 \[ \int e^{\coth ^{-1}(x)} x (1+x) \, dx=-\frac {2 \, {\left (3 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{2}} - 8 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} + 9 \, \sqrt {\frac {x - 1}{x + 1}}\right )}}{3 \, {\left (\frac {3 \, {\left (x - 1\right )}}{x + 1} - \frac {3 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {{\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - 1\right )}} + \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*x*(1+x),x, algorithm="maxima")

[Out]

-2/3*(3*((x - 1)/(x + 1))^(5/2) - 8*((x - 1)/(x + 1))^(3/2) + 9*sqrt((x - 1)/(x + 1)))/(3*(x - 1)/(x + 1) - 3*
(x - 1)^2/(x + 1)^2 + (x - 1)^3/(x + 1)^3 - 1) + log(sqrt((x - 1)/(x + 1)) + 1) - log(sqrt((x - 1)/(x + 1)) -
1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.60 \[ \int e^{\coth ^{-1}(x)} x (1+x) \, dx=\frac {1}{3} \, \sqrt {x^{2} - 1} {\left (x {\left (\frac {x}{\mathrm {sgn}\left (x + 1\right )} + \frac {3}{\mathrm {sgn}\left (x + 1\right )}\right )} + \frac {5}{\mathrm {sgn}\left (x + 1\right )}\right )} - \frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{\mathrm {sgn}\left (x + 1\right )} \]

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*x*(1+x),x, algorithm="giac")

[Out]

1/3*sqrt(x^2 - 1)*(x*(x/sgn(x + 1) + 3/sgn(x + 1)) + 5/sgn(x + 1)) - log(abs(-x + sqrt(x^2 - 1)))/sgn(x + 1)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.95 \[ \int e^{\coth ^{-1}(x)} x (1+x) \, dx=2\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {6\,\sqrt {\frac {x-1}{x+1}}-\frac {16\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{3}+2\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{\frac {3\,\left (x-1\right )}{x+1}-\frac {3\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}+\frac {{\left (x-1\right )}^3}{{\left (x+1\right )}^3}-1} \]

[In]

int((x*(x + 1))/((x - 1)/(x + 1))^(1/2),x)

[Out]

2*atanh(((x - 1)/(x + 1))^(1/2)) - (6*((x - 1)/(x + 1))^(1/2) - (16*((x - 1)/(x + 1))^(3/2))/3 + 2*((x - 1)/(x
+ 1))^(5/2))/((3*(x - 1))/(x + 1) - (3*(x - 1)^2)/(x + 1)^2 + (x - 1)^3/(x + 1)^3 - 1)